Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

 Multiple Choice QuestionsMultiple Choice Questions

1.

A real valued function f(x) satisfies the functional equation f(x – y) = f(x) f(y) – f(a – x) f(a + y) where a is a given constant and f(0) = 1, f(2a – x) is equal to

  • –f(x)

  • f(x)

  • f(a) + f(a – x)

  • f(-x)


A.

–f(x)

f(a – (x – a)) = f(a) f(x – a) – f(0) f(x) 

 = - f(x) [ ∵ x = 0, y= 0, f(0) = f2 (0)-f2(a) = 0 ⇒ f(a) = 0]

146 Views

2.

Suppose f(x) is differentiable x = 1 and limit as straight h rightwards arrow 0 of 1 over straight h space straight f left parenthesis 1 plus straight h right parenthesis space equals space 5 space comma then space straight f apostrophe left parenthesis 1 right parenthesis space equals

  • 3

  • 4

  • 5

  • 6


C.

5

straight f apostrophe left parenthesis 1 right parenthesis space equals space limit as straight h rightwards arrow 0 of space fraction numerator straight f left parenthesis 1 plus straight h right parenthesis minus straight f left parenthesis 1 right parenthesis over denominator straight h end fraction semicolon As the function is differentiable so it is continuous as it is given that limit as straight h rightwards arrow 0 of space fraction numerator straight f left parenthesis 1 plus straight h right parenthesis over denominator straight h end fraction space equals space 5 and hence f(1) = 0 Hence  limit as straight h rightwards arrow 0 of space fraction numerator straight f left parenthesis 1 plus straight h right parenthesis over denominator straight h end fraction space equals space 5
123 Views

3.

Let f and g be differentiable functions satisfying g′(a) = 2, g(a) = b and fog = I (identity function). Then f ′(b) is equal to

  • 1/2

  • 2

  • 2/3

  • None of these


A.

1/2

Given fog = I
⇒ fog(x) = x for all x
⇒ f ′(g(x)) g′(x) = 1 for all x

f ′(g(a)) =1/g('a) = 1/2
⇒f(b') = 1/2

143 Views

4.

The range of the function 7-xPx-3 is

  • {1, 2, 3}

  • {1, 2, 3, 4, 5}

  • {1, 2, 3, 4}

  • {1, 2, 3, 4, 5, 6}


A.

{1, 2, 3}

The given function f(x) = 7-xPx-3 would be defined if
(i) 7 - x > 0 ⇒ x < 7
(ii) x - 3 > 0 ⇒ x > 3
(iii) (x - 3) < (7 - x)
⇒ 2x < 10 ⇒ x < 5
⇒ x = 3, 4, 5
Hence Range of f(x) = {4P0, 3P1, 2P2}
Range of f(x) = {1, 3, 2}

727 Views

5.

Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)} be a relation on the set A = {3, 6, 9, 12} be a relation on the set A = {3, 6, 9, 12}. The relation is

  • reflexive and transitive only

  • reflexive only

  • an equivalence relation

  • reflexive and symmetric only


A.

reflexive and transitive only

Reflexive and transitive only.
e.g. (3, 3), (6, 6), (9, 9), (12, 12) [Reflexive]
       (3, 6), (6, 12), (3, 12) [Transitive].

172 Views

6.

Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}. The relation R is

  • a function

  • reflexive

  • not symmetric

  • transitive


C.

not symmetric

Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} is a relation on the set A = {1, 2, 3, 4}, then
(a) Since ∈ R and (2,3) ∈ R, so R is not a function.
(b) Since (1, 3) ∈ R and (3, 1) ∈ R but (1, 1) ∉ R. So R is not transitive.
(c) Since (2, 3) ∈ R but (3, 2) ∉ R, so R is not symmetric.
(d) Since (4, 4) ∉ R so R is not reflexive,

149 Views

7.

A function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched?

  • Interval Function
    (-∞, ∞) x3 – 3x2 + 3x + 3
  • Interval Function
    [2, ∞) 2x3 – 3x2 – 12x + 6
  • Interval Function
    (-∞, 1/3] 3x2 – 2x + 1
  • Interval Function
    (- ∞, -4] x3 + 6x2 + 6

C.

Interval Function
(-∞, 1/3] 3x2 – 2x + 1

Clearly function f(x) = 3x2 – 2x + 1 is increasing when
f′(x) = 6x – 2 ≥ 0 ⇒ x∈[1/3, ∞)

155 Views

8.

Let f : (-1, 1) → B, be a function defined by straight f left parenthesis straight x right parenthesis space equals space tan to the power of negative 1 end exponent space fraction numerator 2 straight x over denominator 1 minus straight x squared end fraction comma space then f is both one-one and onto when B is the interval

  • open parentheses 0 comma space straight pi over 2 close parentheses
  • [0, π/2)

  • open parentheses negative straight pi over 2 comma straight pi over 2 close parentheses
  • open square brackets negative straight pi over 2 comma straight pi over 2 close square brackets

C.

open parentheses negative straight pi over 2 comma straight pi over 2 close parentheses
Given space straight f left parenthesis straight x right parenthesis space equals space tan to the power of negative 1 end exponent space open parentheses fraction numerator 2 straight x over denominator 1 minus straight x squared end fraction close parentheses space for space straight x element of space left parenthesis negative 1 comma 1 right parenthesis
Clearly space range space of space straight f left parenthesis straight x right parenthesis space equals open parentheses negative straight pi over 2 comma straight pi over 2 close parentheses
therefore space co minus domain space of space function space equals space straight B space equals space open parentheses negative straight pi over 2 comma straight pi over 2 close parentheses
130 Views

9.

The domain of the function straight f left parenthesis straight x right parenthesis space equals space fraction numerator sin to the power of negative 1 end exponent space left parenthesis straight x minus 3 right parenthesis over denominator square root of 9 minus straight x squared end root end fraction

  • [2, 3]

  • [2, 3)

  • [1, 2]

  • [1, 2)


B.

[2, 3)

For the the function straight f left parenthesis straight x right parenthesis space equals space fraction numerator sin to the power of negative 1 end exponent space left parenthesis straight x minus 3 right parenthesis over denominator square root of 9 minus straight x squared end root end fraction space is
we will define

(I) −1≤ (x − 3) ≤1⇒ 2 ≤ x ≤ 4 ....... (i)
(II) 9 x 0 3 x 4 2 − ≥ ⇒ − < < ....... (ii)
From relation (i) and (ii), we get 2 < x < 3
∴ Domain of the given function = [2, 3)

139 Views

10.

The graph of the function y = f(x) is symmetrical about the line x = 2, then

  • f(x + 2)= f(x – 2)

  • f(2 + x) = f(2 – x)

  • f(x) = f(-x)

  • f(x) = - f(-x)


B.

f(2 + x) = f(2 – x)

If y = f (x) is symmetric about the line x = 2 then f(2 + x) = f(2 – x).

219 Views